Overall Equilibrium
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A uniform half metre rule can be balanced at the 29-0 cm mark when a mass 20 g is hung from its one end.
(a) Draw a diagram of the arrangement.
(b) Find the mass of the half metre rule.
A uniform metre rule of mass 100 g is balanced on a fulcrum at mark 40 cm by suspending an unknown mass m at the mark 20 cm.
(i) Find the value of m.
(ii) To which side the rule will tilt if the mass m is moved to the mark 10 cm ?
(iii) What is the resultant moment now ?
(iv) How can it be balanced by another mass of 50 g ?
Define moment of couple. Write its S.I. unit.
What do you mean by clockwise and anti-clockwise moment?
The conditions for equilibrium of a system of parallel forces acting in a plane is/are
algebraic sum of all forces must not be zero.
algebraic sum of moments must be zero.
algebraic sum of all forces must be zero.
algebraic sum of moments must not be zero.
Define the moment of a couple. Write its SI unit.
When a boy weighing 20 kgf sits at one end of a 4 m long see-saw, it gets depressed at this end. How can it be brought to the horizontal position by a man weighing 40 kgf.
Prove that Moment of couple = Force × couple arm.
Explain how a steering wheel rotates with respect to couple.
The diagram in the figure shows a uniform metre rule weighing 100 gf, pivoted at its centre O. Two weights 150 gf and 250 gf hang from the points A and B of the metre rule such that OA = 40 cm and OB = 20 cm. Calculate:
(i) the total anticlockwise moment about O,
(ii) the total clockwise moment about O,
(iii) the difference of anticlockwise and clockwise moments,
and (iv) the distance from O where a 100 gf weight should be placed to balance the metre rule.
You find your mass to be on a weighing machine. Is your mass more or less than ?
What do you understand by the term couple? State its effect. Give two examples in our daily life where couple is applied to turn a body.
A ball is moving over a horizontal smooth surface with a constant velocity. What type of forces is acting on the ball?
The diagram below shows a uniform bar supported at the middle point O. A weight of 40 gf is placed at a distance of 40 cm to the left of the point O. How can you balance the bar with a weight of 80 gf?
What do you mean by equilibrium of a body ?
A body is acted upon by two unequal forces in opposite directions, but not in same line, The effect is that :
(a) the body will have only the rotational motion
(b) the body will have only the translational motion
(c) the body will have neither the rotational motion nor the translational motion
(d) the body will have rotational as well as translational motion.
- the velocity of the body increases.
- net torque is zero but not net force.
- both net force and net torque are zero.
- net force on the body is zero but not net torque.
For rotational equilibrium of a rigid body, it is essential that __
net torque about points lying outside the body must be zero.
All of these
net torque about points lying on the axis of rotation must be zero.
net torque about points lying on the body must be zero.
The diagram in the figure shows a uniform metre rule weighing 100 gf, pivoted at its centre O. Two weights 150 gf and 250 gf hang from the points A and B of the metre rule such that OA = 40 cm and OB = 20 cm. Calculate:
(i) the total anticlockwise moment about O,
(ii) the total clockwise moment about O,
(iii) the difference of anticlockwise and clockwise moments,
and (iv) the distance from O where a 100 gf weight should be placed to balance the metre rule.
When the force applied by the moving vehicle is calculated, the actual force applied by the brakes may be slightly less than the one calculated theoretically. Why?
Due to opposing force of friction
Because action and reaction pair are never equal
Because force of friction is in the direction of motion
Due to mass of the body
- 30 cm
- 90 cm
- 50 cm
- 70 cm
- 1.5 m on the same side as of Raghav
- 1.5 m on the same side as of Raghav
- 1.5 m on the opposite side as of Raghav
- 3 m on the opposite side as of Raghav
- 300 gf
- 50 gf
- 500 gf
- 20 gf
Is the body in rotational equilibrium? But the net torque about p and q is not zero right how? Plz explain thanks.
In the following diagram, the rod is in equilibrium. Then,
F1×d1 < F2×d2
F1×d1 = 3F2×d2
F1×d1 > F2×d2
F1×d1 = F2×d2
Determine the point of application of net force, when forces of 20 N & 30 N are acting on the rod as shown in the figure.
30 cm from
40 cm from
60 cm from
70 cm from point
- 433.33 grams
- 533.33 grams
- 333.33 grams
- 233.33 grams
A fielder pulls his hands gradually with the moving ball while holding a catch. Why does he do this?
It reduces the impact of catching the fast moving ball.
It reduces the frictional forces between hand and the ball
It increases the momentum of the ball.
none
What do you understand by the term couple? State its effect. Give two examples in our daily life where couple is applied to turn a body.